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Is this a standard result with a well known proof? Or would anyone know how to go about proving it?

$$\lim_{k\to\infty}\frac{1}{k}(\sum_{i=1}^k \sin(\omega i)\sin(\omega(i-j))=\frac{1}{2\pi} \int_{0}^{2 \pi}\sin (\omega x)\sin(\omega(x-j))dx,$$ where $j$ is an integer $\ge 0$ and $\omega \in \mathbb{R}$.

I have shown that this seemingly works numerically for large values of $k$, but wanted to see if this is true analytically.

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Using the Riemann sum: $$\lim_{n\to \infty}\sum_{i=1}^nf(x_i)\Delta x=\int_a^bf(x)dx$$ where $\Delta x=\frac{b-a}{n}$ and $x_i=a+\Delta x \cdot i$,

you get, $$\frac{1}{2\pi}\int_0^{2\pi}\sin(\omega x)\sin(\omega(x-j))dx=\lim_{k\to\infty}\frac{1}{k}\sum_{i=1}^{k}\sin\left(\omega\frac{2\pi i}{n}\right)\sin\left(\omega\left(\frac{2\pi i}{n}-j\right)\right)$$ and using trigonometric identities you should be able to show $$\frac{1}{k}\sum_{i=1}^{k}\sin\left(\omega\frac{2\pi i}{n}\right)\sin\left(\omega\left(\frac{2\pi i}{n}-j\right)\right)=\frac{1}{k}\sum_{i=1}^{k}\sin(\omega i )\sin\left(\omega i-j)\right)$$ yet, I was unable to achieve this.

This may be of use: How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?

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  • $\begingroup$ I don't believe this can be directly applied since in the original problem, $i$ are integers from $1$ through $k$, which are not on the interval $[0,2 \pi]$ $\endgroup$
    – Brian Lai
    Apr 2 at 17:17
  • $\begingroup$ @BrianLai what? $\endgroup$
    – Shean
    Apr 2 at 19:43

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